and discharged along C D, the pressure again falling to A. The work
done on the air is the area A B C D. Similarly, in the motor, we admit
along D G, expand from G to H, and discharge along H A, the pressure
rising again to D. The work done by the air is D G H A.
work returned D G H A
work expended A B C D
and these areas are to be measured. For any pressure P, P Vv = C
" * P Vby Cb
and —a = a constant
which is true anywhere on the adiabatics. Imagine the areas cut
into horizontal strips, each of a very small width c^ and let a = area of
one strip to Va, while b is that to Vb. Then a = cVn and I) =
Sum the strips for the whole areas, and let Va == R Vb.
2Vb ^ i_ = Vb
" sVbR "R V"
Area D G H A V
Area A 3 C D Va~
and, = V;
Now suppose ratio of compression ~1 = r
Then by p 607
But, by changing from r2 at c to TJ at G by constant pressure, we
have, by Charles' law:
and ri —